I am in the middle of a homework problem and can't find out what I am doing wrong. The question asks:
Use the transformation in Exercise 2 (given as $u=x+2y$, $v=x-y$) to evaluate the integral $\int_{0}^{\frac{2}{3}}\int_{y}^{2-2y}(x+2y)e^{y-x}dxdy$ by first writing the integral over a region G in the $u v$-plane.
My work is as follows:
Finding x and y in terms of u and v: I found that $x=\frac{u}{3}+\frac{2}{3}v$ and that $y=\frac{u}{3}-\frac{v}{3}$
My work for the above: $v=x-y$ -> $y=x-v$
$v=x+2y$ -> $x=u-2y$
$x=u-2(x-v) = u-2x+2v \equiv 3x=u+2v = \frac{u}{3} + \frac{2}{3}$
$y=u-2y-v\equiv3y=u-v=\frac{u}{3}-\frac{v}{3}$
Then finding the Jacobian using the coefficients of the aforementioned x and y equivalents: \begin{bmatrix} \frac{1}{3} & \frac{2}{3} \\[0.3em] \frac{1}{3} &-\frac{1}{3} \end{bmatrix} $=-\frac{7}{9}$
Solving for the bounds and equation in terms of $u$ and $v$: $\int_{0}^{\frac{2}{3}}\int_{0}^{2}(u)e^{-v}(-\frac{7}{9})dudv$
The answer I get is $-\frac{14}{9}(1-\frac{1}{e^\frac{2}{3}})$, which is not what the answer is supposed to be: $\frac{1}{3}(1+\frac{3}{e^2})$
First of all, your computation of $u$ and $v$ in terms of $x$ and $y$ is correct. However your the determinant of your Jacobian is $- \frac{1}{3}$ so its absolute value is $ \frac{1}{3}$ (Easy to check that) .
Now for your boundaries . In the first integral you have $$ 0 \leq y \leq 2/3 $$ and $$ y \leq x\leq 2-2y$$ so in terms of $u$ and $v$ we get $$ 0 \leq \frac{u-v}{3} \leq 2/3 $$ and $$ \frac{u-v}{3} \leq \frac{u+2v}{3} \leq 2-2\frac{u-v}{3} $$ The first one implies $$ v \leq u \leq 2+v $$ qnd the second implies $$ u-v \leq u+2v \text{ and } u+2v \leq 6-2u+2v$$ and so treating these last two inequalities we get $$ -v \leq 2v \text{ and } u\leq 6-2u $$ i.e. $$ 3v \geq 0 \text{ and } 3u\leq 6 $$ so that $$ v \geq 0 \text{ and } u\leq 2 $$ plotting this region we get
so you integral should be $$\frac{1}{3 } \int_0^2 \int_0^u u e^{-v}dvdu$$
Try to compute it, and you will find the desired result. Hope this helps you.